Two sources of sound are moving in opposite directions with velocities `V_(1)` and `V_(2)(V_(1) gt V_(2))`. Both are moving away from a stationary observer. The frequency of both the source is 1700Hz. What is the value of `(V_(1)-V_(2))` so that the beat frequency observed by the observer is 10Hz. `V_("sound")=340m//s` and assume that `V_(1)` and `V_(2)` both are very much less than `V_("sound")`.

Options

A. `1m//s`
B. `2m//s`
C. `3m//s`
D. `4m//s`

Explanation

To solve the problem, we need to determine the difference in velocities \( V_1 - V_2 \) such that the beat frequency observed by a stationary observer is 10 Hz. We will use the given information and the formula for the frequency heard by the observer.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of both sources, \( f = 1700 \) Hz
- Velocity of sound, \( v_{sound} = 340 \) m/s
- Beat frequency, \( f_b = 10 \) Hz
- Velocities of the sources, \( V_1 \) and \( V_2 \) (with \( V_1 > V_2 \))

2. **Formula for Frequency Heard by the Observer:**
The frequency heard by the observer from each source can be calculated using the Doppler effect formula:
\[ f' = f \cdot \frac{v_{sound} + v_{observer}}{v_{sound} - v_{source}} \]
Since the observer is stationary, \( v_{observer} = 0 \):
- For source 1 (moving away):
\[ f_1 = f \cdot \frac{v_{sound}}{v_{sound} + V_1} = 1700 \cdot \frac{340}{340 + V_1} \]
- For source 2 (moving away):
\[ f_2 = f \cdot \frac{v_{sound}}{v_{sound} + V_2} = 1700 \cdot \frac{340}{340 + V_2} \]

3. **Calculate the Beat Frequency:**
The beat frequency \( f_b \) is given by the difference in frequencies heard from the two sources:
\[ f_b = |f_2 - f_1| = 10 \text{ Hz} \]
Substituting for \( f_1 \) and \( f_2 \):
\[ 10 = 1700 \left( \frac{340}{340 + V_2} - \frac{340}{340 + V_1} \right) \]

4. **Simplifying the Expression:**
Factor out \( 1700 \):
\[ 10 = 1700 \left( \frac{340(340 + V_1) - 340(340 + V_2)}{(340 + V_1)(340 + V_2)} \right) \]
This simplifies to:
\[ 10 = 1700 \cdot \frac{340(V_2 - V_1)}{(340 + V_1)(340 + V_2)} \]

5. **Assuming \( V_1 \) and \( V_2 \) are Much Less than \( v_{sound} \):**
Since \( V_1 \) and \( V_2 \) are much less than 340 m/s, we can use the approximation:
\[ (340 + V_1) \approx 340 \quad \text{and} \quad (340 + V_2) \approx 340 \]
Thus, the equation simplifies to:
\[ 10 = 1700 \cdot \frac{340(V_2 - V_1)}{340^2} \]
Rearranging gives:
\[ 10 = 1700 \cdot \frac{V_2 - V_1}{340} \]

6. **Solving for \( V_1 - V_2 \):**
Rearranging gives:
\[ V_1 - V_2 = \frac{10 \cdot 340}{1700} \]
Simplifying:
\[ V_1 - V_2 = 2 \text{ m/s} \]

### Final Answer:
The value of \( V_1 - V_2 \) is \( 2 \text{ m/s} \).